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4+∑[4*3^n/a^n], n=1 to infinty=16, where a=4

i need to know the numbers outside the circles aswell

f(1)  =   - (1)^2  =  -1

f (f(1) )  =  f (-1)  = -1 + 8  =  7

f (f (f (1) ) )  =  f ( 7)  = - (7)^2  = -49

 f ( f ( f( f(1) ) ) )   =  f (-49)  = -49 + 8  = -41

f ( f ( f ( f ( f ( 1) ) ) ) )  =  f (-41)  =  - (-41)^2  = -1681

From A to C we have all the possible arrangements of this set

( North, North, East)  =  C(3,1)   = 3

From C to B we have all the possible arrangements of this set

( North , North, East, East, East, East )   =  C(6,2)  = 15

So...the total number of paths from  A to B  through C are  3 * 15  = 45

We can answer this with a Venn Diagram :

There are   [ 4 + 10 + 11 + 7 ] = 32 total students

10 play football and cricket....so...the probability that a student plays football and cricket  is

10 / 32   =    5 / 16  =  .3125  =  31.25 %   

s(t)  =  t^3  - 3t

The velocity is given by the derivative of this function  =

s ' (t)   =  3t^2  - 3

Setting the velocity to 0 and solving for t, we have

3t^2  -  3  = 0       factor

3 (t^2 - 1)   = 0        divide through  by 3

t^2  - 1   =0

(t + 1) ( t - 1)  = 0

Setting both factors to 0 and solving for t, we have that

t  = - 1 seconds    or  t  =  1 second

Assuming  that the time is  ≥ 0....its velocity is  0 at 1 second  and its position  is

(1)^3  -  3(1)    =    -2   ⇒  (2 meters to the left of its starting position)

Here are the possible outcomes :

1,1   1,2   1,3   1,4    1,5    1,6

2,1  2,2    2,3   2,4    2,5    2,6

3,1  3,2    3,3   3,4    3,5    3,6

4,1  4,2    4,3   4,4    4,5    4,6

5,1  5,2   5,3    5,4    5,5    5,6

We have 30 possible outcomes

Those which are multiples of 4 and  factors of 15   (in any order) are in red  = 6 outcomes

So....the probability is   6 / 30   =  1 / 5

Note that....on a 12-hour clock....the number opposite  any integer N < 7 is given by :

N  +  (12/2)

And the number opposite any integer N  > 6  is given by :

N - (12/2)

So...this implies that on a 142-hour clock, since 27 is  less than 1/2 of 142, the integer directly opposite 27  is given by :

27  +  (142/2)  =

27  + 71  =

98

In 20 years, it will have

37000 ( 1.05)^20  ≈

98,172  inhabitants

We can choose    1 of any 26 shirts, 1 of any 26 pairs of pants  and 1 of any 26 pairs of shoes...so

C(26,1)  * C(26,1)  * C (26,1)   =

[ C (26, 1) ] ^3  =

26^3   =

17,576 different outfits

Sorry, don't seem to understand your question!!

Here is n with exactly 2 prime divisors: n=5141 = 53 x 97. As you can see it has ONLY 2 prime divisors other than 1 and 5141 itself. If you square it, you always get 9 divisors:

5141^2 =26,429,881 =1 | 53 | 97 | 2809 | 5141 | 9409 | 272473 | 498677 | 26429881 (9 divisors)

So, I don't get how it could have 27 divisors !!!!

"A Senate committee has 8 Republicans and 6 Democrats. In how many ways can we form a subcommittee of 5 members that has at least one member from each party?"

.

6(4x-3) + 7(x+1) = 9

(24x-18) + (7x+7) = 9

31x - 11 = 9

31x = 20

x = 20/31

{22,24,26,28,30,32,34,36,38,40,42,44,46,48} 14 numbers

set a and b already defined, 26 being in both a and b

so there is a 1/14 chance that the number is 26

See

See the answer here:

https://web2.0calc.com/questions/how-many-paths-are-there-from-to-if-every-step-must#r1

I count 48 EVEN divisors:

7! =5,040 = 2 |  4 |  6 |  8 |  10 | 12 | 14 |  16 | 18 | 20 |  24 | 28 | 30 | 36 | 40 | 42 | 48 | 56 | 60 |  70 | 72 | 80 | 84 | 90 | 112 | 120 | 126 | 140 | 144 | 168 | 180 | 210 | 240 | 252 | 280 | 336 | 360 | 420 | 504 | 560 | 630 | 720 | 840 | 1008 | 1260 | 1680 | 2520 | 5040 (48 divisors)

The probability that a student chosen at random has a cat and a dog   =   9 / 25

4 sin²x   =   4 cos x  +  5

                                                   By the Pythagorean identity,   sin²x  =  1 - cos²x    so....

4( 1 - cos²x )   =   4 cos x  +  5

                                                   Distribute the  4  to the terms in parenthesees.

4  -  4 cos²x   =   4 cos x  +  5

                                                   Add   4 cos² x   to both sides of the equation.

4   =   4 cos²x  +  4 cos x  +  5

                                                   Subtract  4  from both sides of the equation.

0   =   4 cos²x  +  4 cos x  +  1

                                                   Split   4 cos x   into two terms that we can use to factor by grouping.

0   =   4 cos²x  +  2 cos x  +  2 cos x  +  1

0   =   2 cos x ( 2 cos x  +  1 )  +  1( 2 cos x  +  1 )

0   =   ( 2 cos x  +  1 )( 2 cos x  +  1 )

0   =   ( 2 cos x  +  1 )²

                                      Take the square root of both sides.

0   =   2 cos x  +  1

                                      Subtract  1  from both sides.

-1   =   2 cos x

                                      Divide both sides by  2 .

-1/2   =   cos x

The angle that has a cosine of   -1/2   in the interval  [0, pi)   is   2 pi / 3

x  =  2 pi / 3

You are right! Was too hasty!. 3^3 x 5^3, 3^3 x 5^6, 3^3 x 5^9, 3^6 x 5^3, 3^6 x 5^6, 3^6 x 5^9=

3375, 421875, 52734375, 91125, 11390625, 1423828125

I think I got them all: 5 + 6 = 11 perfect cubes? Since you know what to do, you can check them and see if I missed any.

I don't think thats right because 5^3*3^3 is also a cube.

3^6*5^{10}

=7119140625 = 3^6 * 5^10

From factoring above, you will have: 3^3, 3^6, 5^3, 5^6, 5^9 =27,  729, 125, 15625, 1953125=5 perfect cubes.